Question

What is the complete factorization of the polynomial below

Answer 1

Answer:

$a. \ (x-2)(x+i)(x-i)$

Step-by-step explanation:

First of all let's notice that $(x-i)(x-i)=(x-i)^2 = x^2+2xi-1 \in \mathbb{C}$ while our original polynomial is real. So we can rule out the even options B and D. At this point, if the polynomial has a factor of $(x-\alpha)$ it means tha $\alpha$ is a zero of the polynomial. Let's check both 2 (for option a) and -2 (for option c)

$p(2)=2^3-2(2)^2+2-2= 8-8+2-2=0$

$p(-2)= (-2)^3-2(-2)^2+(-2)-2=-8-8-2-2=-20$

At this point 2 is a zero, and our final factoring is a$(x-2)(x+i)(x-i)$